# Math Help - Torus and smooth function

1. ## Torus and smooth function

Hello,

I want to show that this function is smooth:

$
f:T->\mathbb{R} , f(x)=e^1(x)=x^1
$

T is the Torus defined as: $T=\{((R+r*cosv)cosu,(R+r*cosv)sinu,r*sinv) \subset \mathbb{R}^3: u,v \in [0,2\pi)\}$ and R,r>0 constant.

I don't know how f could be differentiated. My Problem is, that f is not in the standard form f(x,y,z)=x. it is something like f((R+r.....))=(R+r*cosu)cosv.
I'm a little bit confused. Is here any trick or something

regards

2. in your case you just need to show that f(u,v) = (R + r cosv)cosu is smooth, since the parameterization (u,v) is already smooth.

3. Hello,

i have shown, that f(x,y,z)=x is smooth (the projection on first coord.) and we know that our parametrization is also smooth, i.e. $\phi(u,v)=(R+r*cos.....,...,..)$
But i'm not really sure, why this is enough for being smooth related to f.

Anyway:

Now i want to show, that our function f is a morse function.
But i can't see the critical points of f??

if i differentiate f(x,y,z)=x i get Df=(1,0,0): $\mathbb{R}^3->\mathbb{R}$ this is a surjective function isn't it? so there are no critical points.

Can you explain me, what is wrong in my way of thinking?

Thanks

Regards

4. a "smooth structure" needs to be defined to define smooth functions, just as a manifold does. So I said "in your case", that is, if we don't involve the abstract definition of a manifold, we just define a "smooth structure" on a surface as its smooth parametrization, that is, a smooth map from an area D of R^2 to R^3.
Then a smooth function from a surface is defined to be the smooth function from its parameter domain D.
For your new question, f DOES have critical points on T. You can see this from the expression f(u,v) = (R + r cosv)cosu, or just think that f is defined on a compact surface it must have maximum and minimum points, those points being critical points.
The problem of your way is, you need to "restrict" f to T. Suppose the natural inclusion map from T to R^3 is i, that is, i(x,y,z)=(x,y,z), then the gradient of f restricted to T is i*(df), that is, the pull back of df by i.